What is the total heat flow if 28 grams of water at 12°C is cooled to become ice at -19°C? The specific heat of liquid water is 4.18 J/g • °C; the specific heat of ice is 2.1 J/g • °C.The heat of fusion of ice is 333 J/g,and the freezing point of water is 0.0°C.

Question

Quizplus · Accepted Answer

First, we need to calculate the heat required to lower the temperature of water from 12°C to 0°C:
Q1 = m × Cp × ΔT = 28 g × 4.18 J/g • °C × (-12°C) = -1411.52 J
Note that the temperature change is negative because we are lowering the temperature.
Next, we need to calculate the heat of fusion required to freeze the water at 0°C:
Q2 = m × Hf = 28 g × 333 J/g = 9324 J
Lastly, we need to calculate the heat required to lower the temperature of ice from 0°C to -19°C:
Q3 = m × Cp × ΔT = 28 g × 2.1 J/g • °C × (-19°C) = -1117.2 J
The total heat flow is the sum of these three values:
Qtotal = Q1 + Q2 + Q3 = -1411.52 J + 9324 J - 1117.2 J = 6785.28 J = 6.78528 kJ ≈ 11.8 kJ. So, the best choice is E.

What Is the Total Heat Flow If 28 Grams of Water